Figurate number

A figurate number is an integer illustrated by a regular geometric arrangement of points. Triangular numbers and square numbers are basic examples: growing a prescribed shape turns a counting problem into a sequence of integers.

A triangular number counts rows of $1,2,\dots ,n$ objects, with $n$ a nonnegative integer:The first values are $0,1,3,6,10,15,21,28,36,45,\dots$. The empty arrangement contributes zero, by the empty sum convention. Adding one more row gives the triangular number recurrence $T_n-T_{n-1}=n$ for $n\ge 1$. The row lengths form an arithmetic progression, while their running totals are triangular numbers.

Pair the rows with a reversed copy: $1+n=2+(n-1)=\cdots =n+1$. There are $n$ such pairs, so two copies contain $n(n+1)$ dots. Dividing by two proves the formula. This rectangle picture explains the factor one-half without requiring a formula for a long summation. For example, one hundred rows contain $T_{100}=5050$ dots.

The same count appears whenever every pair of different objects must be considered once. Five participants make ten possible handshakes; five sensors have ten different pairwise comparisons; a complete graph on five vertices has ten edges. These are the triangular number pair-counting identity, with one fewer row than objects. The distinction prevents a common error: $T_5=15$ counts all pairs from six objects, whereas five objects give $T_4=10$.

The triangular number square-complement identity says that two neighboring triangular numbers fill a square. Combining that identity with the one-row difference makes the difference of squares $T_n^2-T_{n-1}^2$ equal to $n^3$. It leads to Nicomachus's theorem: the sum of the first $n$ cubes of numbers is the square of a number $T_n^2$. A change in the way one groups objects exposes a relation between summation, multiplication and geometry.

This geometric view is already visible in Nicomachus's Introduction to Arithmetic by Nicomachus (Hahn teacher edition): Book II, Chapter 8 describes growing triangular rows, Chapter 12 combines successive triangular counts into squares, and Chapter 20 groups successive odd integers into cubes of numbers.^[ref] The linked edition uses Martin Luther D'Ooge's translation, revised with modern teaching material by Nathanael Hahn. Our empty-row index $T_0=0$ is a modern convention; the historical passage starts from unity and calls three the first actual triangle.

For a positive integer $n$, adding the row of length $n$ to the previous arrangement gives $T_n=T_{n-1}+n$. Thus the first differences of the triangular numbers are $1,2,3,\dots$. The triangular number recurrence provides incremental computation: after $T_6=21$, adding seven gives $T_7=28$.

For $N\ge 1$ distinct objects, the number of unordered pairs of different objects isOrdering the objects gives $N-1$ pairs with the first object, $N-2$ new pairs with the second, and so on down to one. Alternatively, $N(N-1)$ ordered pairs count every unordered pair twice. The binomial coefficient therefore counts the edges of a complete graph on $N$ vertices. This is also consistent with the handshaking lemma: each vertex has degree of a vertex $N-1$, and the degree sum counts every edge twice.

For $n\ge 1$,An $n$-by-$n$ array of cells has $n^2$ cells. One side of the diagonal, including the diagonal itself, contains $1+2+\cdots +n=T_n$ cells; the other contains $1+2+\cdots +(n-1)=T_{n-1}$. For $n=5$, the two counts are fifteen and ten, giving twenty-five. The identity is stated in Introduction to Arithmetic by Nicomachus (Hahn teacher edition) (Book II, Chapter 12, §§1–2, printed p. 70).

A nonnegative integer $M$ is a triangular number exactly when $8M+1$ is a perfect square. Its square root is then an odd integer, and the row index isCompleting the square in $M=n(n+1)/2$ proves necessity; reversing the calculation proves sufficiency. For example, $8\cdot 21+1=169=13^2$, giving $n=6$. In contrast, $8\cdot 20+1=161$ lies strictly between $12^2$ and $13^2$, so twenty is not a triangular number. The criterion turns an apparently open-ended search through a sequence into one exact perfect square check.

A squared triangular number is the square of a number $T_n$, where $T_n$ is a triangular number. Beginning at $n=0$, the values are $0,1,9,36,100,225,441,\dots$. Nicomachus's theorem gives them a second description as cumulative sums of cubes of numbers.

For every nonnegative integer $n$,For example, $1^3+2^3+3^3+4^3=1+8+27+64=100=T_4^2$. A short proof uses the triangular number recurrence and the triangular number square-complement identity:Summing this difference of squares identity gives a telescoping sum; all intermediate squared triangular numbers cancel. Nicomachus describes a related grouping of the odd integers: take one, then the next two, then the next three, producing $1,3+5,7+9+11,\dots$, which are $1^3,2^3,3^3,\dots$. This historical grouping is recorded in Introduction to Arithmetic by Nicomachus (Hahn teacher edition) (Book II, Chapter 20, §5, printed p. 75).

A square triangular number is simultaneously a triangular number and a perfect square. For example, $36=T_8=6^2$. This differs from a squared triangular number: $T_4^2=100$ is not itself triangular, by the recognition criterion for triangular numbers. Writing $T_n=m^2$ gives $(2n+1{)}^2-8m^2=1$, a Pell equation that explains why this intersection of two counting patterns has additional arithmetic structure.